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%I #27 Nov 07 2019 09:03:34
%S 1,3,13,17,1073,19,17,29,10000117,73,17,3,1007,3,43,11,
%T 1000000000000029,1,31,11,1191,1,1143,31,10000079,21,91,59,1019,3,67,
%U 117,10000000000000000000000000000077,109,89,49,1097,41,1053,43,10000047,87,23,53,1149,83,57
%N Smallest k such that k//A002275(n)//k is prime, where // denotes concatenation and A002275(n) is the n-th repunit (R_n).
%C a(n) = 3 if n is in A056251.
%C From _Chai Wah Wu_, Nov 05 2019 : (Start)
%C Theorem: a(2^r*s) >= 10^(2^r-1) for all r >= 0, s > 0.
%C Proof: Note that if k has m digits, then k//A002275(n)//k = k*(10^(n+m)+1) + A002275(n)*10^m which is a multiple of gcd(A002275(n),10^(n+m)+1).
%C Next, since 10^(2^r) - 1 = (10^(2^(r-1) -1))*(10^(2^(r-1) + 1)) and 9 does not divide 10^n+1, by induction it is easy to see that 10^(2^w) + 1 is a divisor of A002275(2^r) for 1 <= w < r. Since A002275(2^r) is a divisor of A002275(2^r*s), 10^(2^w) + 1 is also a divisor of A002275(2^r*s).
%C For 1 <= m < 2^r, let t be the 2-adic valuation of m, i.e. 0 <= t = A007814(m) < r.
%C Then 10^(2^r*s+m)+1 = 10^(2^t*q)+1 = (10^(2^t))^q + 1 for some odd number q.
%C Since the sum of two odd powers a^q+b^q is divisible by a+b, this implies that 10^(2^r*s+m)+1 is divisible by 10^(2^t)+1.
%C This means that for n = 2^r*s and 1 <= m < 2^r, gcd(A002275(n),10^(n+m)+1) >= 10^(2^t)+1 > 1, i.e. k//A002275(n)//k is not prime.
%C Thus a(2^r*s) must have at least 2^r digits, i.e. a(2^r*s) >= 10^(2^r-1). QED
%C As a consequence, a(n) >= 10^(A006519(n)-1). This result is still true if some of the digits of k are leading zeros.
%C (End)
%H Chai Wah Wu, <a href="/A263182/b263182.txt">Table of n, a(n) for n = 0..1023</a>
%F a(A004023(n)-2) = 1. - _Chai Wah Wu_, Nov 04 2019
%e R_0 = 0 and the smallest k such that k//0//k is prime is 1, so a(0) = 1.
%t Table[k = 1; While[! PrimeQ[f[n, k]], k++]; k, {n, 0, 7}] (* _Michael De Vlieger_, Oct 13 2015 *)
%o (PARI) a(n) = my(rep=(10^n-1)/9, k=1); while(!ispseudoprime(eval(Str(k, rep, k))), k++); k
%Y Cf. A002275, A004023, A006519, A056251.
%K nonn,base
%O 0,2
%A _Felix Fröhlich_, Oct 11 2015
%E a(16)-a(46) from _Chai Wah Wu_, Nov 04 2019
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